A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillation

Question

A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of $24\ \mu	ext{T}$. When a horizontal field of $18\ \mu	ext{T}$ is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be:

Accepted Answer

1. **Time Period Formula:** The time period ($T$) of a magnet oscillating in a uniform magnetic field ($B$) is given by $T = 2\pi \sqrt{\frac{I}{mB}}$, where $I$ is the moment of inertia and $m$ is the magnetic moment. This implies $T \propto \frac{1}{\sqrt{B}}$.
2. **Initial Condition:**
   - Initial magnetic field ($B_1$) = Earth's horizontal field ($H$) = $24\ \mu	ext{T}$.
   - Initial time period ($T_1$) = 2 s.
3. **Final Condition:**
   - An external field ($B_{ext}$) of $18\ \mu	ext{T}$ is produced *opposite* to the Earth's field.
   - Net magnetic field ($B_2$) = $H - B_{ext} = 24\ \mu	ext{T} - 18\ \mu	ext{T} = 6\ \mu	ext{T}$.
4. **Calculation:**
   - Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$:
     $$ \frac{T_2}{2} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2 $$
     $$ T_2 = 2 	imes 2 = 4 	ext{ s} $$

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